A 2.5 kg box is held released from rest 1.5 meters above the ground and slides d

Question

A 2.5 kg box is held released from rest 1.5 meters above the ground and slides down a
frictionless ramp. It slides across a floor that is frictionless, except for a small section 0.5 meters
wide that has a coefficient of kinetic friction of 0.4. At the left end, is a spring with spring
constant 250 N/m. The box compresses the spring, and is accelerated back to the right.
What is the speed of the box at the bottom of the ramp?
What is the maximum distance the spring is compressed by the box?
What is the maximum height to which the box returns on the ramp?

Explanation / Answer

energy conservation

potential energy = energy used against friction and compression energy in spring

mgh= u*mg*d + 0.5*k*x*x

2.5*1.5*9.8 = 0.4*2.5*9.8*0.5 + 0.5*250*x*x

spring compression = x = compression = 0.5047 m

now to find speed before friction surface at end of ramp

potential energy = kinetic energy

mgh=0.5*m*v*v

2.5*9.8*15=0.5*2.5*v*v

V at bottom of ramp = 5.422 m/s

again

now

compression energy in spring = energy used against friction + potential energy

0.5*k*x*x = u*m*g*d + m*g*h

0.5*250 *x*x = 0.4*2.5*9.8*0.5 + 2.5*9.8*h

h = 1.096 m

hope i explained upto u ask,please give rating

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